Answer:
[tex]y=3x^{2} -18x+15[/tex] is the equation the quadratic as per the given conditions.
Step-by-step explanation:
Let the equation of the quadratic be:
[tex]y=ax^{2} +bx+c[/tex]
Given that it has zeroes at x = 1 and x = 5 i.e. y = 0 at both the values of x.
Also passes through point (3, -12). i.e. when x = 3, y = -12
Putting x = 1, y = 0:
[tex]\Rightarrow a\times 1^{2} +b\times 1+c=0\\\Rightarrow a+b+c=0 ....... (1)[/tex]
Putting x = 5, y = 0:
[tex]\Rightarrow a\times 5^{2} +b\times 5+c=0\\\Rightarrow 25a+5b+c=0 ....... (2)[/tex]
Putting x = 3, y = -12:
[tex]\Rightarrow a\times 3^{2} +b\times 3+c=-12\\\Rightarrow 9a+3b+c=-12 ....... (3)[/tex]
Equation (2) - Equation (1):
[tex]24a+4b=0\\\Rightarrow 6a +b=0 ...... (4)[/tex]
Equation (2) - Equation (3):
[tex]16a+2b=12\\\Rightarrow 8a +b=6 ...... (5)[/tex]
Equation (5) - Equation (4):
[tex]2a=6\\\Rightarrow a =3[/tex]
Putting value of a in equation (4):
[tex]6\times 3+b=0\\\Rightarrow b = -18[/tex]
Putting a and b in equation (1):
[tex]3-18+c=0\\\Rightarrow c= 15[/tex]
So, the quadratic equation is:
[tex]y=3x^{2} -18x+15[/tex]
